Problem: What is the slope of the line tangent to $f(x) = x^{2}-x-8$ at $x = 2$ ?
Solution: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{((x+h)^{2}-(x+h)-8) - (x^{2}-x-8)}{h}$ $ = \lim_{h \to 0} \frac{(x^{2}+2x h+h^{2}-(x+h)-8) - (x^{2}-x-8)}{h}$ $ = \lim_{h \to 0} \frac{x^{2}+2(x h)+h^{2}-x-h-8-x^{2}+x+8}{h}$ $ = \lim_{h \to 0} \frac{2(x h)+h^{2}-h}{h}$ $ = \lim_{h \to 0} 2x+h-1$ $ = 2x-1$ $ = (2)(2)-1$ $ = 3$